# Inverting Consequences of Uniform Convergence

1. Jan 17, 2010

### r.a.c.

Hi. Now you probably know that if a function fk(x) converges uniformly to f(x) then we are allowed to certain actions such as

$$limn-> \inf \int f (of k) dx = \int f dx$$

In other words we are allowed to exchange limit and integral. Now say we have any sequnce valued function fk(x) . And we also have

$$limn-> \inf \int f (of k) dx = 0$$

Does that imply that fk(x) converges uniformly to 0?

2. Jan 17, 2010

### Landau

Indeed, if $$(f_k)_{k\in\mathbb{N}}$$ converges uniformly to $$f$$ on some interval $$[a,b]\subset\mathbb{R}$$, then $$\lim_{k\to\infty}\int_a^b f_k(x)dx=\int_a^b \lim_{k\to\infty}f_k(x)dx=\int_a^b f(x)dx$$.

The converse of this is:

if $$\lim_{k\to\infty}\int_a^b f_k(x)dx=\int_a^b f(x)dx$$, does it follow that $$(f_k)_{k\in\mathbb{N}}$$ converges uniformly to $$f$$?

This is not what you asked, after all $$\int_a^b f(x)dx=0$$ does not imply $$f=0$$ (i.e. f(x)=0 for all x).

The converse is therefore obviously not true. Take for example [a,b]=[0,1], and define f_k and f on [0,1] by f_k(x)=1/k and f(x)=-1 if x<1/2, f(x)=+1 if x>1/2. Then

$$\lim_{k\to\infty}\int_0^1 f_k(x)dx=\lim_{k\to\infty}\frac{1}{k}=0=-\frac{1}{2}+\frac{1}{2}=\int_0^1 f(x)dx$$

but (f_k)_k does not even converge to f pointwise.

For an counter-example to your claim, just take (f_k)_k to be a constant sequence: f_k=f as defined above, for all k. Then the integral of f_k is equal to zero (and hence the limit of the integral too), but f_k does not converge (even pointwise) to the zero function.

Last edited: Jan 17, 2010
3. Jan 17, 2010

### r.a.c.

Thanks a lot. And will do.

4. Jan 17, 2010

### Landau

You're welcome. You can just click on any equation to see its Latex code.